Question

1. Evaluate x+1 using Mathematica. I. (4+1+y)d4; R DA; R = {(x,y): 0 5xs1, 2s y s3} and check your answer

Answer 1

$\therefore \int \int_R (\frac{1}{x+1}+y) dA = \int_{x=0}^1 \int_{y=2}^3 (\frac{1}{x+1}+y) dx dy$

$= \int_{x=0}^1 [(\frac{y}{x+1}+\frac{y^2}{2})]_2^3 \ dx$

$= \int_{0}^1 [(\frac{3}{x+1}+\frac{3^2}{2})-(\frac{2}{x+1}+\frac{2^2}{2})] \ dx$

$= \int_{0}^1 [(\frac{3-2}{x+1}+\frac{9-4}{2})] \ dx$

$= \int_{0}^1 [(\frac{1}{x+1}+\frac{5}{2})] \ dx$

$= [(log(x+1)+\frac{5x}{2})]_0^1$

$= (log(1+1)+\frac{5\times 1}{2})-(log(0+1)+\frac{5\times 0}{2})$

$= log(2)+\frac{5}{2}-log(1)-0$

also, $log(1)=0$

$\therefore \int \int_R (\frac{1}{x+1}+y) dA = log(2)+\frac{5}{2}$

using Mathematica also, we get the same result. ( screenshot is attached below).

In[1]= Integrate[1/(x+1)+y, {x, 0, 1}, {y, 2, 3}] 5 Out[1]= - +Log[2] 2